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Câu hỏi: Cho hàm số bậc bốn $y=f\left( x \right)$. Biết rằng hàm số $g\left( x \right)=\ln f\left( x \right)$ có bảng biến thiên như sau:
Diện tích hình phẳng giới hạn bởi các đường $y={f}'\left( x \right)$ và $y={g}'\left( x \right)$ thuộc khoảng nào dưới đây?
A. $\left( 38;39 \right)$.
B. $\left( 25;26 \right)$.
C. $\left( 28;29 \right)$.
D. $\left( 35;36 \right)$.
A. $\left( 38;39 \right)$.
B. $\left( 25;26 \right)$.
C. $\left( 28;29 \right)$.
D. $\left( 35;36 \right)$.
Ta có: $g\left( x \right)=\ln f\left( x \right)\Rightarrow {g}'\left( x \right)=\dfrac{{f}'\left( x \right)}{f\left( x \right)}$. Nên: ${g}'\left( x \right)=0\Leftrightarrow {f}'\left( x \right)=0$.
Mà ${g}'\left( {{x}_{1}} \right)={g}'\left( {{x}_{2}} \right)={g}'\left( {{x}_{3}} \right)=0\Rightarrow {f}'\left( {{x}_{1}} \right)={f}'\left( {{x}_{2}} \right)={f}'\left( {{x}_{3}} \right)=0$.
Theo giả thiết bài toán thì $f\left( x \right)>0\forall x$ nên: ${g}'\left( x \right)>0\Leftrightarrow {f}'\left( x \right)>0\Leftrightarrow \left[ \begin{aligned}
& {{x}_{1}}<x<{{x}_{2}} \\
& x>{{x}_{3}} \\
\end{aligned} \right.$
Và: $f\left( {{x}_{1}} \right)=10,f\left( {{x}_{2}} \right)=42,f\left( {{x}_{3}} \right)=37\Rightarrow {g}'\left( x \right)-{f}'\left( x \right)={f}'\left( x \right)\left( \dfrac{1}{f\left( x \right)}-1 \right)>0\Leftrightarrow \left[ \begin{aligned}
& {{x}_{2}}<x<{{x}_{3}} \\
& x<{{x}_{1}} \\
\end{aligned} \right.$
Vậy:
$S=\int\limits_{{{x}_{1}}}^{{{x}_{3}}}{\left| {g}'\left( x \right)-{f}'\left( x \right) \right|}dx=\int\limits_{{{x}_{1}}}^{{{x}_{2}}}{\left( {f}'\left( x \right)-{g}'\left( x \right) \right)dx}+\int\limits_{{{x}_{2}}}^{{{x}_{3}}}{\left( {g}'\left( x \right)-{f}'\left( x \right) \right)dx}$
$\text{ }=\int\limits_{{{x}_{1}}}^{{{x}_{2}}}{{f}'\left( x \right)\left( 1-\dfrac{1}{f\left( x \right)} \right)dx+\int\limits_{{{x}_{2}}}^{{{x}_{3}}}{{f}'\left( x \right)\left( \dfrac{1}{f\left( x \right)}-1 \right)dx}}$
$\text{ }=\int\limits_{{{x}_{1}}}^{{{x}_{2}}}{\left( 1-\dfrac{1}{f\left( x \right)} \right)d\left( f\left( x \right) \right)+\int\limits_{{{x}_{2}}}^{{{x}_{3}}}{\left( \dfrac{1}{f\left( x \right)}-1 \right)d\left( f\left( x \right) \right)}}$
$\begin{aligned}
& \text{ }=\left. \left( f\left( x \right)-\ln \left| f\left( x \right) \right| \right) \right|_{{{x}_{1}}}^{{{x}_{2}}}+\left. \left( \ln \left| f\left( x \right) \right|-f\left( x \right) \right) \right|_{{{x}_{2}}}^{{{x}_{3}}} \\
& \text{ }=27+\ln 10-\ln 37 \\
& \text{ }\approx 25,69 \\
\end{aligned}$.
Mà ${g}'\left( {{x}_{1}} \right)={g}'\left( {{x}_{2}} \right)={g}'\left( {{x}_{3}} \right)=0\Rightarrow {f}'\left( {{x}_{1}} \right)={f}'\left( {{x}_{2}} \right)={f}'\left( {{x}_{3}} \right)=0$.
Theo giả thiết bài toán thì $f\left( x \right)>0\forall x$ nên: ${g}'\left( x \right)>0\Leftrightarrow {f}'\left( x \right)>0\Leftrightarrow \left[ \begin{aligned}
& {{x}_{1}}<x<{{x}_{2}} \\
& x>{{x}_{3}} \\
\end{aligned} \right.$
Và: $f\left( {{x}_{1}} \right)=10,f\left( {{x}_{2}} \right)=42,f\left( {{x}_{3}} \right)=37\Rightarrow {g}'\left( x \right)-{f}'\left( x \right)={f}'\left( x \right)\left( \dfrac{1}{f\left( x \right)}-1 \right)>0\Leftrightarrow \left[ \begin{aligned}
& {{x}_{2}}<x<{{x}_{3}} \\
& x<{{x}_{1}} \\
\end{aligned} \right.$
Vậy:
$S=\int\limits_{{{x}_{1}}}^{{{x}_{3}}}{\left| {g}'\left( x \right)-{f}'\left( x \right) \right|}dx=\int\limits_{{{x}_{1}}}^{{{x}_{2}}}{\left( {f}'\left( x \right)-{g}'\left( x \right) \right)dx}+\int\limits_{{{x}_{2}}}^{{{x}_{3}}}{\left( {g}'\left( x \right)-{f}'\left( x \right) \right)dx}$
$\text{ }=\int\limits_{{{x}_{1}}}^{{{x}_{2}}}{{f}'\left( x \right)\left( 1-\dfrac{1}{f\left( x \right)} \right)dx+\int\limits_{{{x}_{2}}}^{{{x}_{3}}}{{f}'\left( x \right)\left( \dfrac{1}{f\left( x \right)}-1 \right)dx}}$
$\text{ }=\int\limits_{{{x}_{1}}}^{{{x}_{2}}}{\left( 1-\dfrac{1}{f\left( x \right)} \right)d\left( f\left( x \right) \right)+\int\limits_{{{x}_{2}}}^{{{x}_{3}}}{\left( \dfrac{1}{f\left( x \right)}-1 \right)d\left( f\left( x \right) \right)}}$
$\begin{aligned}
& \text{ }=\left. \left( f\left( x \right)-\ln \left| f\left( x \right) \right| \right) \right|_{{{x}_{1}}}^{{{x}_{2}}}+\left. \left( \ln \left| f\left( x \right) \right|-f\left( x \right) \right) \right|_{{{x}_{2}}}^{{{x}_{3}}} \\
& \text{ }=27+\ln 10-\ln 37 \\
& \text{ }\approx 25,69 \\
\end{aligned}$.
Đáp án B.